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What are ways to reduce drag and noise generated by the landing gear?
Apart from using fairings, what ways are there, or have been studied, but not necessarily implemented in reducing drag and noise produced by the landing gear, mainly for commercial aircraft with retractable landing gear, but would also like to know about smaller aircraft, perhaps with non-retractable landing gear.
Is there a study or a design that has been done on landing gear to reduce drag and/or noise produced by the landing gear?

You should choose between drag and aerodynamic noise, and the aircraft type you are interresting in. Moreover, you should show your current state of research (e.g. do you see the dr 400 landing gear?).

@ManuH The main aim of the research is to reduce noise, but it seems that there would be the benefit of drag reduction to follow. As mentioned the aircraft type is a commercial airliner, specifically the A350-900. See my comment to the answer below for information on the current state of research.

The biggest noise reduction is achieved by closing most of the landing gear doors again after extension is complete.
Of the factors that influence drag and noise, one is the distance between wheel and fuselage. The picture below is taken from S. Hoerner's book "Fluid Dynamic Drag", chapter XIII.5 of which deals with the drag of landing gears.

Another influence is the shape of the landing gear strut. Since most airliner landing gear legs are oleo struts, they are round for practical reasons. Fairing them would already help, as can be seen from the lower line in the plot above. Another improvement can be achieved by covering the wheel hubs. However, the reduction of the drag and noise caused by extended, retractable landing gears has been studied but few of the results have yet found their way into actual aircraft. According to this DLR report, a fully faired A320 landing gear would reduce noise by 10 dB while partially fairing the most noisy parts of an A340 landing gear would reduce noise by 5 dB. The picture below is taken from this report.

For fixed landing gears, I recommend NACA Technical Report 485 which contains results of full-scale tests on several gears. NACA Technical Report 518 contains results for partially retractable gears.

That is exactly what I have been looking into. I am actually trying to retrofit a strut fairing to the main landing gear of the A350-900. The main aim is to reduce noise, but it seems obvious that there would also be the benefit of drag reduction.
With that being said, I am currently looking into and trying to figure out the constraints there are to work with, and what exactly is the area or space available to be used to retrofit said fairing. It would have to not restrict any of the mechanisms and take advantage of any space available.

I don't know that non-retractable landing gear make any noise.  If they do, it's certainly not audible above the engine.
To reduce drag, fairings are the only method I know of.  I am currently replaceing my old beat up 1973 vintage ABS plastic fairings with newer more aerodynamically shaped fairings to reduce drag. All 3 wheels, main gear legs, and also around the nose scissor jack. And going to a more aerodynamically shaped tail cone. All should contribute to a speed increase, and will look sharper after being painted.  You can see them here:
http://www.aircraftspeedmods.ca/products.htm
If you look at the Red Bull racing airplanes, you can see similarly shaped fairings, and what looks to be much smaller wheels as well (single seat planes vs a 4 seater). Take a look AT Extra 300S airplanes, I think that is what they are using now.

Aeroengineer Roy Sobchuk did tests with tufts of string to determine where air was most turbulent to help come up with the shapes used on my model of airplane.

Looks like Red Bull Air Race pilots fly several types of planes http://airrace.redbull.com/en_US/article/planes

Of course they make noise. This adds up, and especially with engine idling during approach the landing gear generates a considerable part of the overall noise of airplanes.

What kind of plane are you referring to? In my fixed gear single airplane, they aren't making much noise.

@PeterKämpf All sources of aircraft noise, and the latest technology in reducing aircraft noise are summed up in this DLR presentation.

Well, a lot of that doesn't apply to my Cessna Cardinal https://en.wikipedia.org/wiki/Cessna_177_Cardinal and the OP did ask "would also like to know about smaller aircraft, perhaps with non-retractable landing gear."

I just came across  a neat NASA video that 'shows' noise around a large jet landing gear. https://www.youtube.com/watch?v=zd5kmdUzgdQ


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Would there be a purpose for a set of compressors in an electric turbo fan?
This is actually a two part question. Would there be a purpose to a compressor in an electric turbofan where there is no need for a combustion chamber, and if not, is there then a purpose for a bypass as well?
Sorry if I'm unclear, as i am not an industry professional perhaps my wording may not be clear. What I mean to say is that in a regular turbofan engine there are compressors that push air into a combustion chamber with higher density (presumably to get more potential energy per volume of air) where fuel is introduced and ignited to create higher pressure/kinetic energy to run the turbines,  which in turn move the front compressors and the fan. However, in an electrically driven fan, since there is no combustion happening, and the fan is driven by an electric motor, does the engine then need the compressors? Do they perform any benefit in a system that does not use combustion to create pressure?
Then if not, is there any benefit to the bypass system in an electrically driven fan engine? Or does it make more sense to simply switch to a ducted fan system?

unclear what you are asking.

Sorry if I'm unclear, as i am not an industry professional perhaps my wording may not be clear.

The "rear compressors" are called turbines and are not compressors at all since they actually decompress the flow; you can look up this terminology with a cursory Wikipedia search.

yes thank you for the correction.

No.  The only purpose of the compressor is to provide high pressure air for the combustion chamber and turbine.  If there is no combustion chamber and no turbine, then you do not need a compressor.  Just drive a big ducted fan (or propellor, your choice) with your electric motor.

Yes, if you want to go super sonic.
This effectively gives you the performance characteristics of a low-bypass turbofan or a turbojet:
1) low cross section and high thrust to cross section ratio (low supersonic drag)
2) achieving very high nozzle velocity by jetting the high pressure flow out of the nozzle (useful at high speed)
3) Flow increases pressure but not velocity when moving through the engine, so that all stages operates in the subsonic region. Important if your intake is not far from transonic when the airplane is supersonic.
4) Poor economy.
In the perceivable future it's hard to imagine any jet engine not powered by a turbine (piston ICE, electric, nuclear, etc, any shaft engine except a turbine).
Especially the focus now on electric airplanes is to improve range, and both supersonic and jet kills the range, so practically the answer is still "No".

so basically if I was looking at an engine for subsonic, there's no benefit to having either a compressor, or a by-pass system? Just stick with a normal ducted fan or prop?

@No One . Yes. Just like turbine counterparts, if a prop can do, then prop.

There won't be an electric turbofan as it doesn't make sense to make one. A jet engine is designed to burn fuel and turn it into propulsion, compressors squash air so it can be mixed with fuel. If you take the fuel away and replace it with an electric motor an engine would be designed along completely different principles.

It would be just a fan or propeller.

@TimNevins.  Except....electric turbofans are being made. See http://www.eurekamagazine.co.uk/design-engineering-news/airbus-siemens-and-rolls-royce-to-fly-hybrid-electric-turbofan-in-2020/165547/.  The graphic in the story shows a fan driven by a 2kW motor, exactly as described in the question. And, several other demonstrator aircraft with electric engines are also planned. Cheers.

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What is the metal bit in the front of this propeller spinner?
What is that grey spirally bit in front of the propeller mounted on this Shvetsov ASh-62 radial engine?

The aircraft it's connected to seems to be a Lisunov LI-2, a licensed version of the DC-3 built by Russia. This may be the only flying example of the aircraft.

It is a fitting for a Hucks Starter.
Photos from Vintage Wings Canada

I saw the first photo and assumed the starter shaft was just connected to some typical-looking airport equipment. The second photo was quite the surprise,.

Here I was hoping it was a bottle opener... :/

I'm getting a picture of somebody forgetting to chock the aircraft wheels, and the starter truck crashing into something while reversing out of the way at top speed :)

@FreeMan It's a Russian engine. The vodka is in the fuel tank, not in the nose cone.

Both the airplane and the truck are beautiful.  Just beautiful.

I had a 1964 Datsun pickup that had a fitting like that -- it fit the truck's jack handle, but it was in fact for starting the engine.  Actually used it once...

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Why don't airports use arresting gears to recover energy from landing passenger planes?
An average plane has 50t of mass and a landing speed of approx. 250 (lets be conservative) 200 km/h. Thus its kinetic energy at landing is about 75 MJ or 21 kWh.
The plane catches the wire attached to a pneumatic or hydraulic system, which will decelerate the plane and harvest its kinetic energy. The built up pressure in the system can be released via some kind of turbine to convert the pressure (potential energy) to electricity. Lets say the system has 50% efficiency. Leading to 10kWh harvested energy, per landing.
The airport in Hamburg, Germany is just average. Not large and not small. They had 150 000 flight movements in 2017. Lets say half of them are big planes and 1/2 were landings. That is 37,5k landings per year. Lets say 2/3 could use the arresting gear system:
This would provide a cumulative 250 MWh of harvested electricity per year. This is equivalent to the electric energy consumption of 100 two person households in Germany.
In Germany 1kWh electricity costs about 30 euro cents (including tax etc). An airport could save 75000 euro per year when harvesting this energy and use it, instead of buying it from the net.
I can't imagine, that such a harvesting system would not pay down after a couple of years.
Why there are no systems like this in the aviation industry?
Con:

You need to refit thousands of planes with a wire catching System
System has to be safe in case of a go-around
???

EUR75000 is negligible compared to the development, certification and operation costs of such a system. Look into the development costs of carrier arresting systems for an idea of the difficulties. Also, 50% energy recovery sounds extremely optimistic, unless you have a source to support that assumption (?).

I'd pay extra to not get the "fighter jet landing experience" as a passenger. Not to mention it's a lot more than just putting a hook on, the whole airframe would have to be more robust. You'd probably spend much more in added fuel costs than you'd ever harvest...

The landing could be as smooth as it is now for normal landing. I do not want to make the plane brake immediatly.

Even if it was, you have training issues as well. Naval pilots can hit an exact spot every time because the runway can move to minimize cross-wind. Airliners landing on airports don't get that luxury of wind always coming down the runway. This means the landing spot can change, a lot. Then you have the issue of running over wires, replacing the wires (on carriers, they are replaced every 125 landings). I just don't see this being economical or practical.

There's also the power grid side.  Injecting impulse power like that would play merry hell with your system power flow & stability - just the opposite of the brownouts you get when e.g. starting a large electric motor.  So you'd have to add something like flywheels or capacitors that could store the energy...

Naval pilots don't hit the exact spot every time vertically, when they are landing in a 10 meter sea swell. The peak accelerations on the aircraft during carrier landings are often of the order of 10 to 30g - I once was involved in measuring them, to find why bits of the aircraft kept breaking unexpectedly. Carrier landings are basically "controlled crashes."

It would be cheaper (for the airport) to insist that planes that are parked should run their engines on idle and use the excess power to put electricity into the local grid. As an added bonus, the retrofit costs would be minimal.

@alephzero: 30g?  So tell us, just how do they keep the pilots' eyeballs from popping out?

@jamesqf - I would assume the forces are not distributed evenly into the pilot's compartment. That's what suspension is for.

A business like an airport does not really feel 75'000 EUR. I think. That is approximately the equivalent of 1-2 persons working there. Furthermore such device, I agree is useful to reduce power usage, but might fail, should be repaired, so that means one has to hire people to fix it (eventually).

I'd remark that a much more plausible implementation of such a system would be electric motor/generators in the landing gear itself, charging batteries in the plane. The main benefit wouldn't be the recovered energy on landing though, but the ability to use less engine power before takeoff. Still I'm pretty sure the weight of the batteries would eat up the fuel savings – unless they can also be used for ascend.

If you think that something that earns 75k per year will pay for itself in a couple of years, then you think that thing will cost about 150k. That's peanuts.

@Richard; So the plane is experiencing 30g vertical deceleration (from hitting the deck), not a 30g horizontal deceleration?

From your figures, it appears to work out to about 2 Euros per landing. An aircraft will easily burn up more than that just idling for a minute or so. An airport that can manage traffic well to minimize time spent queuing for take off or waiting for vacant terminal gates would save airlines huge amounts in fuel costs - saving they would probably not mind sharing with the airport - far more than a couple of EU per landing.

Consider revising your design by replacing runways with reverse conveyor belts that will be put in motion by the friction of landing aircraft. With those you can land in either direction, don't need to land in the exact spot, and won't need to retrofit aircraft.

Even if it was doable, just have one single accident with at least one passenger killed, which has at least some chance to be connected to this system... and say goodbye to this system, all the investment costs + the costs of lawsuits.

On aircraft carriers, what happens to the energy "eaten" by the catching system? Is it just dissipated as heat? If so, that seems to be a lot of heat. Could tehy use that energy, even if just  for heating the captain's coffee?

@HagenvonEitzen - that's grounds (pun intended!) for a whole new question. Please feel free to ask it!

The energy unit costs for companies in Germany are much lower than for households. Where as a citizen you pay approximately 30 ct / kWh, the airport will pay much less, probably around 15 ct / kWh.
Adjusting your calculation for this price, we have 37.5 k€ per year in terms of energy cost saving to offset the depreciation and maintenance of the installation.
Let's assume our bean-counters want to see their investment money back in 10 years (10% depreciation per year), and the yearly maintenance costs are 10% of the initial system price. That means the cost of such an installation at Hamburg airport can be at most 187500.
But mind you, Hamburg has two runways, and each can be used in two directions. To catch 2/3 of the total traffic, we would need to install at least two systems. Runway 23 and runway 15 combined get approximately 75% of the landing traffic, they would be the preferred choice. (traffic percentages per runway are here)
This brings the maximal cost of a single installation to 95 k€; including project management costs, costs of the equipment, cost of the installation works, cost of closing the runway etc etc.
That is absolutely nothing! You can't develop and install a system that needs to recouperate 21 kWh in thirty seconds (the duration of the deceleration phase) for such a price.
And then there is the airborne side of the story. Who is going to pay for the modifications needed on the aircraft, such as an arrestor hook? And for the fuel that it costs to carry this hook around during flight (a hint, it will cost more energy to carry this extra weight around all the time, than that it would harvest at the airport)? And who is paying for the certification of the system?
There is no business case for such a system, not even close.

Just look at the 'savings' per flight.

Not only business (capitalism is a bad guidance), it is likely physically ineffective, therefore environmentally ineffective?

I imagine one would use an aircraft carrier style sling, except much longer (have fun managing the unrolling - and then rolling up - of about 2km of tape) and softer (so that the cargo doesn't come out damaged). Still, if the pilot catches the sling and then decides for touch-and-go... the recuperator is likely to break and cause significant damage to airport equipment.

Maybe if you could replace 95 KEUR with 95k €? (If I am guessing correctly that by K you meant k, a prefix for kilo=1000, not K=1024.)

Also, 21kWh generated in a few seconds would need to be stored somewhere - the grid isn't just a dumping ground where you can pour electricity in like a pot whenever you like.  This system would need huge infrastructure to store this generated power and deliver it to the grid at a sustained regular rate.

@J... Actually, I think the grid can very comfortably absorb 75MJ (unless you have references to show it can't).

If that's 75MJ in 15 seconds (planes don't stop immediately after all), you only have 5MW. It doesn't need to be absorbed by the grid at all; that's locally consumed by the airport. IOW, no need to upgrade the grid connection of the airport. Having said that, the engineering solution would be a flywheel that's spun up by the unspooling of the cable. That already gives you the energy store needed.

@MartinBonner Grid connection follows rules, but each case is also evaluated individually (connection impact assessment) - what may be acceptable in one area may not be in another depending on the capacity of the local grid. The grid certainly does not like spikes, however, so I doubt they would approve a direct-to-grid momentary dump generation scheme like this.  You can't magically increase the grid load so all it would do is generate a voltage spike.  The energy producer would almost certainly be required to regulate the output to deliver it continuously over a much longer period of time.

@MSalters Yes, a flywheel would probably be part of the solution, but you could not connect it directly to the arresting cable.  The latter would need special engineering to match the variable load profiles required by various aircraft, I think, and the main point is that it would certainly add to the (already prohibitive) cost of the system.

You can't simply stick an arrestor hook on a passenger airplane, the fuselage is not designed for the forces that a rear hook would impart. Carrier airplanes are designed from the beginning for carrier landings and have beefed up structures. That extra structure adds weight to them (presuming it's even possible to do so), and it would also add weight to the passenger airplanes as well, making your airplanes less efficient and using more fuel - more than any possible savings you'd get from your proposed system.

Carrier aircraft must land on a "runway" that is about 100m long.  The OP is not proposing to stop aircraft this quickly.  Modifications would be needed, for sure, but not to AC levels.

Yes, dynos work at many speeds - presumably such a system would be engineered to decelerate the craft at its normal rate otherwise generated by the brakes.  This isn't the main reason not to implement such a system - if it were otherwise profitable and sensible this requirement could easily be accommodated.

@LawnmowerMan making the cables much longer (like in kilometers which is what airliners need to stop) is impractical at best. Cables are in use for emergencies at some military fields, and they're not significantly longer than those at carriers. They're only used in emergencies as there's almost guaranteed damage to the aircraft and pilot when they are used.

Of course the hook doesn't need to be attached to the fuselage directly. It can be integrated in the landing gear. The landing gear can take the forces

The drag on a cable several hundred meters long is probably more than the power you can recover from the device. Plus, because you drag that cable over the runway, you have to replace it regularly, adding to the cost again.

@DeltaLima, the hook needs to be in line with the center of the mass of the aircraft or it will pull the aircraft to one side or the other, I don't see how you could integrate the hook with the landing gear.

Summary: It isn't worth it financially, quite apart from any other aspects:

The 'savings' would be "lost in the noise" of operating an airline service- they are absolutely minuscle compared to all other costs involved.
The amortised cost of implementation + operation + maintenance + safety management would greatly exceed the savings.

________________________________________
Using your own figures, the savings amount to about 3 Euros per flight. That's less than the cost of an airline meal, or a passenger purchased drink or almost any identifiable discretionary customer related service item.
Even before the complexities and costs of development deployment and use are considered, the available income is far far less than could be achieved by action with truly minimal low impact low development and low deployment costs within the airline passenger service structure.

Not to mention how much the cost to an airline can vary depending on how much time an individual aircraft spends in a holding pattern or even just idling on the ground, waiting for one thing or another.

Beyond the other issues mentioned here, let's accept your 75,000 euro/year number for a second and ignore the impossibility of retrofitting aircraft. How many people does it take to operate the arresting gear system? This article says it takes seven people for a US Navy aircraft carrier. But let's say your system is very efficient, someone will work very long hours, and it somehow takes only one person to operate the system, retracting the arresting wire between aircraft, maintaining the equipment and generator, etc... That person will want to be paid. How much does it cost to keep him or her employed? Seems like a pretty skilled position, got to be at least 75,000 euro a year, right? Without even paying for the equipment or any of the other parts of the system, your cost savings have already disappeared.
Also consider that arresting gear cables have a quite limited lifespan and are replaced frequently during use, frequently enough to require replacement every few hours based on the number of operations at a busy commercial airport. You could easily spend more than 75,000 euro/year just on cable replacements. Runway downtime while cables are being replaced may cause aircraft to burn more fuel while they wait; the cost of that fuel could easily exceed the value of the electricity generated by the system.
For another way of looking at it, even if we use your electricity price, consider that 75,000 euro a year is not much compared to the 269 million euro turnover of Hamburg Airport last year. While I'm sure the airport would like to save money where it can, it's clear that retrofitting thousands of planes (if that even was possible, which it's not) to try to save 0.03% of annual revenue would not be practical.
You could also look at the amount of electricity generated. "The electric energy consumption of 100 two person households in Germany" sounds like a lot of electricity, but it's also around the same amount of electricity that you'd get if you could run a single jet engine at takeoff power for a couple hours. Downrate that to a power setting you can maintain continuously, and your whole system is still only producing in a year what a single jet engine outputs in half a day.

There's also a major safety issue involved - when those cables snap it can be catastrophic to any ground crew nearby.  There's a reason that carrier landing crew is among the most dangerous jobs in the military.

One more nail in the proverbial coffin...
Let's say for a second that the airport could install a system like this at zero cost, not entirely unlike the assumption in Zach Lipton's answer.
Let's also say for a second that there are no ongoing costs for the airlines in having their airplanes retrofitted to be able to use such a system. In other words, the arresting hook has zero mass and causes zero net drag increase.
Let's also say for a second that the ground system requires no maintenance and no additional manpower. In other words, the cost for the airport after the system is installed for free is zero.
Let's also say for a second that the electricity can be fed into the ground systems with no losses.
In such a completely unrealistic scenario, now ask yourself who gets the benefit?
The primary benefit here is that airport saves some electricity. Specifically, they need 21 kWh less grid electricity per landing of a reasonably large jetliner. At your €0.30/kWh, that works out to €6.30.
That reasonably large jetliner might have 200 passengers on board.
In other words, a savings of €0.0315 per passenger.
Sure there are people who shop around for low fares, but a €0.03 difference per passenger per flight? And that's assuming that the airport pays full price for the electricity in the first place, and that the airport passes the full electricity cost savings on to the airlines, and that the airline passes the full landing fee savings on to the passengers.
I'll bet you almost anything that there are ways to save €0.03 per passenger per flight that don't require huge, maintenance-demanding, costly ground systems along with adding both mass, drag and certification headaches (and costs) to the airframes. I wouldn't be surprised if something silly like filing a millimeter off one side of each tray table would save more than that from lower fuel consumption because of hauling around less mass throughout the flight.

Setting aside the costs and complexities of such a system, it's also important that the energy you can recover this way is limited to the aircraft's kinetic energy at touchdown. This is a small fraction of the total energy obtained from the energy stored in the aircraft's fuel.
For instance, the Boeing 737NG series, which lands in the 50t ballpark of your example, has about a 20t fuel capacity. Aviation fuel has about a 44 MJ/kg energy density, so a 737NG can get through up to about 800 GJ of energy in one flight, or about 100,000 times the estimated 75 MJ available to an arresting gear.
The vast majority of this consumption is directly related to the aircraft's total mass. The majority of fuel is burned during cruise, when engine thrust predominantly balances drag, about 30% of which is lift-induced drag that is directly proportional to the cruising weight of the aircraft. Other components, like take-off thrust, predominantly increase in more direct proportion to the aircraft mass, so 30% is probably a reasonably conservative estimate of the marginal increase in fuel burn for additional aircraft mass.
It therefore only takes an increase in mass of about 1/30,000 for the energy recoverable to be cancelled out completely by additional fuel burned during a flight. For a 50t plane we can therefore only afford to add a couple of kilograms at most to stand any chance of saving any energy at the end of a flight at the upper end of the aircraft's operating range.
Suffice to say the gear necessary to transfer 75 MJ to systems on the ground in the space of a few seconds will not be this lightweight.

To put things in perspective, 75000€ is the amount an airline will pay in compensations and duties of care on a single flight with ~100 passengers arriving late. So, if a single flight per year is delayed because of the landing system failure in the receiving airport, or the airplane getting damaged after such a landing, you'll instantly lose all the profits. A second incident will make it a net negative.
And I'm not even talking about yearly amortization / service / repair / insurance costs which will be a couple of order of magnitude higher.

No need to modify any airplanes.
You would make the first half of the runway just past the touchdown strips into  a conveyor belt. The conveyor belt uses radar to determine the aircraft's ground speed, and expends energy to spin up the conveyor to projected speed of the airplane on touchdown.  The airplane touches down, and its wheels never spin up, it just sits.
Now, a braking mechanism on the conveyor belt brakes the the belt to a halt, such that the airplane will be at 0 mph speed (0 kph) by the end of the conveyor belt, recovering all the energy in theory. *Mind you, it is also recovering the energy it expended putting the belt into motion in the first place.
When the airplane is down to about 30 mph/50 kph (or whatever its comfortable fast-exit speed is), the pilot releases the brakes altogether, and is now independent of the conveyor, i.e. Its mass is now disconnected from the conveyor's and its own inertia carries it forward independent of whatever the conveyer may do.
The airplane runs straight off the end of the conveyor and finishes its stop on plain pavement, and/or takes a high speed exit at that point.
The conveyor must failsafe by stopping hard instead of freewheeling.  If that happens, the airplane's wheels will suddenly start turning again, and the ordinary braking they are already applying will be effective.   Would they be locked up and skid? Probably not. Very likely the ABS will be modulating braking force so the wheels barely turn with the conveyor's normal slowing; now that they're turning again it will reapply.
Handling might be non-intuitive if the airplane tried to exit in the middle of the conveyor belt.  Simply don't provide any exits there.  If the pilot needs to throw it into the grass at some point, the handling oddity will be the least of his problems.
Of course, this requires essentially frictionless bearings  and slide surfaces, and all this for under €187k capitalization.
And when you realize that is futile, visit your local commuter train station and watch the trains pull in.  Trains weigh a LOT more than airplanes.

You could make your answer much more concise by keeping just the first and the last sentence and deleting the rest :-). Also, many trains already have all the necessary hardware needed to implement regenerative braking, yet it's still often not worth it.

regen braking is wonderful in electric cars and in underground trains -- the latter because the train produces (much) less waste heat.

I'm pretty sure there's an XKCD about airplanes on conveyor belts. As for the trains, that is the whole point. Go watch trains in a modern country, which has electrified tracks. As the train decelerates, the energy is returned to the overhead wire. Even diesel locomotives turn speed back into electricity, only to turn that into heat. It's just better manageable that way.

I find this proposal actually quite intriguing. Advantages: 1. no need anymore to turn into the runway direction last-second, when landing in strong crosswind 2. reduced tire wear – no spin-up skidding at touchdown 3. no brake wear 4. conveyor belt motors can be used for accelerating starting planes. — Too bad the belts would probably cost many millions, and have enourmous bearing etc. losses, and wear out quickly themselves.

@leftaroundabout I think of it as a fun high school science project in the age when we have cold fusion and replicators, and can make this stuff out of nanofibers and buckyballs with lunch money.  Of course, at that point, every single airport will be condo complexes and malls, except for Mirabel, and of course Schonenfeld/Brandenberg which will be scheduled to replace Tegel next year...

@MSalters oh, that's harder than you think.  BART didn't make any friends at PG&E with their idea to regen onto the grid.  (heretofore, regen was pushed out into the rest of the system at great transmission loss, as somewhere a train was surely motoring). Sacramento is playing with a battery substation, that would be sublime for absorbing regen if you had enough of them). Als re: "modern country", sad fact is, America is just too big.  Electrification makes sense Bergen-Oslo.  Spokane-Minneapolis, different deal.

If a plane needs to go around after touchdown (eg: runway incursion), but its already caught on the arrester system, that is one bad scenario.
If the arrester system fails after being caught, then the pilot has an extremely brief window to switch to traditional braking (reverse thrust, control surfaces and brakes).  That is another really bad scenario.
Landing is one of the most dangerous and complex phases of flight.  This portion of a flight should be about safety, not saving a few dollars.  An energy re-capture system increases complexity and decreases safety.

So what you are suggesting is to replace a well-tested process for stopping planes, self contained and redundant (brakes) by a complex mechanism which fundamentally takes away control from the captain (the go-around, as all other maneuvers is the decision of the captain) or has a super complex control handover between plane and ground systems. Not to mention that the time which the system has to recover from stopping a plane may be short or even negative (the time between the plane leaving the runway and the next plane being signaled by the tower to land). You even would have to introduce some additional indicator in the cockpit which requires a last second decision from the pilot in case  the hook is not ready.
(and yes, I know they do it on aircraft carriers but for much better reasons that saving 3 euro per landing).


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What is the typical value of maximum lift coefficient for aerobatic aircraft?
I'm working on a conceptual design phase project for aerobatic aircraft (for university) and I can't seem to find any statistics about key aerodynamic properties (such as in the question), so I hope somebody here does know something about this subject?

Related but not identical

If it's NACA 0012/0015 it will be in Abbott & Von Doenhoff, Theory Of Wing Sections.

First, don't take the two-dimensional value from an airfoil chart and use it for the full aircraft. Subtract 15-20% for wingtip margin and fuselage influence. Wingtip margin means that the wing is less loaded towards the tips to keep a lift margin for aileron control and to avoid a tip stall.
Next, aerobatic aircraft have lower maximum lift coefficients than GA aircraft. Their wing loading is lower, so they can afford this. A symmetric airfoil mounted at zero incidence has proven to be the best choice, and while some older aerobatic aircraft use cambered airfoils, the camber creates a distinct disadvantage in transitions from normal to inverted flight and vice versa.
And last, the maximum lift coefficient depends on quite a list of parameters. If you need a value for the calculation of structural loads, make sure you add enough strength to cover the increased lift possible with high pitch rates.
If you need a figure for a single-seater piston-powered aerobatic aircraft without flaps for the estimation of minimum speed, use a lift coefficient of 1.3.

Actually, engine thrust angle and the fuselage angle create a lifting body, add to the lift component and is considered to cancel the asymmetric span lift. So using a CL based on wing area alone comes close enough to real world results to be commonly used for theoretical performance figures. The one adjustment usually made is increased drag due to tip vertices and aspect ratio. An elliptical wing is considered 100% efficient while a rectangular wing is about %90. Using the mean chord width also negates changes in Reynold number for a tapered wing. https://en.wikipedia.org/wiki/Elliptical_wing

@jwzumwalt: May I politely disagree? Please read this answer for my take on the efficiency of elliptic wings.

It looks to me like you are attempting to refute real world wind tunnel tests by NASA (Previously NACA) Technical Note 2249 and reproduced on pg 75 of the "Aerodynamics for Naval Aviators" with personal philosophy. I'll go with the wind tunnel tests...

There are many factors including tip vortex, prop wash, the wing root is usually a MUCH THICKER profile than wingtip for structural, fuel, and retractable wheels and gives higher Reynold number and lift (higher Reynold number can make as much as 10% difference for the exact same airfoil), wing washout which decreases angle of attack and lift towards the tip, etc. As I mentioned below all these effects cancel each other so the wind tunnel CL data works just right when the entire wing area is used to calculate; it is a very close proven & dependable rule of thumb :)

The primary textbook for many (possibly most) University aerodynamic classes is FREE (Aerodynamics for Naval Aviators). This book explains all theoretical and practical knowledge needed to have a well rounded understanding of aerodynamics. It explains all the formulas needed to design an airplane in the real world. It does not get into the bizarre out of touch academia :)
Page 75 discusses span lift.
https://www.faa.gov/regulations_policies/handbooks_manuals/aviation/media/00-80T-80.pdf
To directly answer your questions regarding aerobatic CL.  Most purely refined aerobatic airplanes will use a symmetric or modified symmetric airfoil. The exact coefficient of lift depends on shape of the leading edge, chord width, and Reynold number (~speed vs chord width).
The 1.3 given above would be close to typical - perhaps a little low, but it depends on how rounded the leading edge is and the design speed of the aircraft. Airfoil books I have show very thin sharp symmetric airfoils at 1.2 and very fat ones at 1.6. Most are 1.4 to 1.5. But remember, the exact same airfoil will show minor differences due to wind tunnel peculiarities, and now most "testing" uses computer fluid dynamic simulation!
An example of a popular aerobatic plane is Extra 300 (symmetric airfoil) often used in Red Bull Competition https://en.wikipedia.org/wiki/Extra_EA-300
You can study various CL at
http://airfoiltools.com/
Other good books I use
https://www.scribd.com/document/322901165/the-science-of-flight-w-n-hubin-1992
https://www.scribd.com/document/322676208/design-for-flying-david-b-thurston-1978
https://www.scribd.com/document/24571577/aircraft-flight-testing-design
An excellent basic series of how2 design aircraft was published in the EAA "Sport Aviation" magazine from February 1990 thru February 1991 primarily by John Roncz.

Airfoil data is not the same as wing data.

...although for this wing with no washout and constant airfoil, 3D data would be pretty close to 2D. Just tip losses and wing/body interference to account for. +1, I like the navy book.

Except for service ceiling, we are generally most concerned with stall and maxCL while landing. It turns out that all factors including ground effect pretty much cancel each other out and an aircraft will stall while landing so close to the projected wind tunnel data for an airfoil that seldom any other computation is needed. The standard wind tunnel test section is 3ft long and the resultant data is mathematically adjusted/predicted for full scale applications.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How does outflow valve work?
As we know that outflow valve normally closes during climb and cabin altitude also increase but why does outflow valve open if we have selected manual pressurisation and increasing cabin altitude !

What else would you expect it to do when manually commanding a higher cabin altitude (i.e. one more like the outside pressure)?

The engines provide air to pressurize the cabin.  This pressure increases as the engines provide air into the cabin.  The outflow valve opens to relieve pressure in the cabin.  When they are in equilibrium, the pressure changes stabilize.  A higher pressure will reduce cabin altitude while a lower pressure will increase cabin.  Hence opening the outflow valve reduces pressure and the result is a higher cabin altitude.

Thank you guys .

If you're selecting increasing cabin altitude, you are lowering cabin pressure.  To do that you have to let some air out of the balloon (the pressure hull).  So the valve has to be opened more (they usually never fully close because there is always more inflow than needed to maintain cabin pressure with both engines running - how much open depending on how leaky the pressure hull is).

In normal operation, the Outflow Valve does not generally completely close except maybe for system self test or validation test. It maintains some flow so as to achieve ventilation of the cabin.
The basic system of operation of outflow valves is to restrict the airflow through the valve so as to maintain an 'artificially' higher pressure of air in the pressurized parts of the aircraft. High pressure air is typically pumped into the cabin by the pneumatic system from engine bleeds after some processing to make the pressure and temperature reasonable for use. The difference between the free air pressure outside the aircraft and pressure inside the cabin is called Differential Pressure and is typically around 7 to 8 psi at cruising levels. This translates into a cabin altitude of 6 to 8 thousand feet in terms of the Std Atmosphere.
In fact, a closed outflow valve, in many aircraft, denotes a major air leak due to a rupture somewhere in the pressurized part of the cabin and is a sign of structural damage.
System diagrams displaying the actual valve position may show it as almost closed for most of the flight during operation in Automatic Mode. In manual mode, the pilot has to adjust the valve position immediately upon selecting 'Manual' to avoid an uncomfortable change in cabin altitude.

6 to 8 thousand feet, I presume?


text:
why is rhumb line bearing for this question 270 as the answer?
An aircraft leaves position "G" in lat 40 degree South and flies the following Rhumb Line tracks and distances:
G to H : 180 (true) 240 NM
H to J : 270 (true) 240 NM
J to K : 000 (true) 240 NM

what is the rhumb line bearing from G to K ??
According to me answer should be 090 but it is given as 270 ..
can someone please give the reason for the same

The only movement in the east-west direction is your 2nd leg, moving west. Therefore, from the starting point G to the ending point K, the bearing is west, 270 degrees. The answer 090 would describe the course from K to G to close the path.
K--->090--->G
^           |
|           V
J<---270<---H

All the legs are the same length, and the north/south aren't drawn to the same scale as the east/west, but it's just a square.  270 from H to J, so 090 from K to G.  And thus 270 from G to K.

The bearing is correct, but the legs are not all the same length because the surface of the Earth is curved. The distance K <-> G will be larger than the other legs since the question said we are south of the equator.


text:
Does shrouding a propeller minimize induced drag by equalizing the downwash velocity along its blades?
EDIT: It's not a duplicate of Are ducted fans more efficient? That question and the answers doesn't address the reason for the higher theoretical efficiency, it is more about efficiency in practice (drag on the duct, weight, etc) and hence why they aren't used despite the higher theoretical efficiency
I was told shrouded propellers are more efficient becuase tip vorticies are eliminated by the wall which would imply no induced drag but apparently that is wrong Do ducted fans eliminate induced drag? therefore I've been trying to figure out why they are still more efficient than an open propeller despite still having induced drag. It must have less induced drag.
The vortex around an unshrouded propeller:

At first I thought the wall somehow increases the effective wingspan, moving the vorticies to the top of the wall much like winglets do and thus reducing the induced drag that way

However unlike winglets the walls don't have a pressure difference on either side (it's not an airfoil) therefore the vortex can't be there.
So the vortex must be around the whole wall because above the wall the pressure is low and below it the pressure is high.

But this doesn't change the effective wingspan so why does it have less induced drag?
My explanation is that the existence of the wall causes the inside of the vortex to "straighten out" with the flow through the duct which makes the downwash velocity constant along the length of the blades (since the flow is irrotational). This means that this condition is satisfied:

from page 7 of http://naca.central.cranfield.ac.uk/reports/1923/naca-report-121.pdf
Therefore induced drag is minimized. Is this correct?

@CrossRoads In this question I clearly say that I thought the explanation was what I wrote in that other question (lower induced drag due to increased effective wingspan) but now I think it's something else (lower induced drag due to more elliptical lift distribution)

@CrossRoads        H      O      V      E      R

The approximate speed that a ducted fan is no longer efficient is about 100mph. So it depends on the specific flight conditions and aircraft configuration.

@jwzumwalt Not doubting your statement, but is there any information citing the 100mph efficiency limit? I would be interested in learning about the physics behind this.

Interest peaked on ducted fans in the late 70's and early 80's. I got my information from the EAA "Sport Aviation" magazine (Aug 68, Jun 73, etc) which ran several authoritative articles that included engineers such as Molt Taylor. He considered it for the Aero Car and Mini-IMP but found it impractical. Some articles included wind tunnel testing and drag analysis. It shouldn't be to hard to find with a Google Search. Peter Kemp on this forum is one of the more prominent aerodynamist and would also be a good source - he loves math and probably has it somewhere in his library.

@CrossRoads I'm asking about induced drag, the drag due to lift. Mentioning possible parasitic drag on the duct wall from forward flight is irrelevant even if this wasn't about a craft in hover.

@fooot That answer just says "yes they are in theory but there are some practical reasons why they aren't used" This discussion is about exactly why they are theoretically more efficient

Basically, yes. The difference between shrouded and unshrouded propeller is that the shrouded one can produce uniform thrust across the diameter, while for the unshrouded one the thrust decreases near the tips.
That way a shrouded propeller accelerates more air than an unshrouded one of the same diameter. This air therefore needs to be accelerated to lower speed, and therefore carries away less kinetic energy, requiring less induced power¹.
However, diameter can be varied, so the efficiency comparison is not that straightforward. When the propeller spins relatively slowly, making it larger is better, similarly to how increasing wing span is better, aerodynamically, than adding winglets.
However increasing the speed of the tips increases parasite drag, especially if it becomes supersonic. And since increasing size while maintaining angular speed increases the orbital speed of the tips, increasing size only helps to a certain point. That's when shrouds become useful.

¹ In propellers and rotors it is called induced power rather than induced drag, because it counts directly against the engine power. It also describes the physics better, since in both cases it is the work that is done on the air by the reaction to the generated lift/thrust.

@Сократ, please, be so kind and ask a question. Comments are not suitable for this kind of discussion.

I suppose the confusion may be due to different sources using different definitions of induced drag. As used most commonly in my experience induced drag is the energy wasted as non-useful work in the creation of turbulence directly attributable to the generation of lift. Most noticeably in the wingtip vortex. In other words imparting flow in directions other than the ideal direction; in keeping with the photo-quote, an infinitely long completely uniform wing would have on lateral pressure gradient and thus no sidewards flow, nor the vortex  formed from sideways flow.(It is not the vortex that causes drag, the vortex is just a symptom of the lateral pressure gradient.)
A close shroud, possibly even attached to propeller tips, of sufficient width will stop these vortex. However it may not stop all lateral flow because of the helical flow of the propwash. The helical flow can be reduced with static blades much like those used in axial flow compressors on gas-turbine engines. The shroud does increase add parasitic and form drag but it most definitely does reduce induced drag.
The first drawing in the Question with the un-shrouded prop, has the wrong perspective or axis so the vortex is in the wrong spacial plane; the second drawing has the vortex flowing through a solid wall and the prop again has the wrong axis. The third drawing is not a vortex it is the bulk flow encountered with a stationary fan in an enclosed room.(and the prop has the wrong rotational axis)

Induced drag has only one definition: the work done on the air to produce lift. Most of this work is imparting the flow the desired direction, and is therefore useful. You correctly note that the vortex is just a symptom, not cause, but it is mainly result of the lift generation itself, not the sideways flow. Induced drag is mainly proportional to lift to span ratio. An infinite wing only has no induced drag if you only let it produce finite total drag, which means zero lift per unit of span. If you let it produce non-zero lift per unit of span, it will have induced drag too.

Induced drag is a term used well beyond aircraft design, so I'm only using "lift" as an example familiar to the aviation field but other forms of dynamic fluid reaction could be applied. A wing with no sideways flow produces no vortex yet still produces lift. Wingspan ratio reduces induced drag by reducing the lateral pressure gradient and by extension induces less lateral/sideways flow.

I suppose I could narrow the term to three possible definitions: 1, total drag created from the creation of useful lift plus the energy wasted on non-ideal flow as a result of lift; 2, just the portion of drag attributed to the energy wasted on non-ideal flow as a result of lift; 3, only the portion of drag attributable to lift produced by an ideal-flow.

Everybody except you agrees that induced drag is what you have under option 1.

No, the vortices are trapped in the tip clearance.
What if there's no "outside" at all to your setup? Imagine a theoretical scenario where the entire space outside the duct is solid to $\infty$. Where are the vortices now? They could only be within the tip clearance.
I just realized this paper from another question is the perfect answer to this question.
And your assumption of shroud somehow making the downwash uniform is also wrong. Note that although the drawing in the statement of the problem or this paper is for a two-blade shrouded fan, even a fan with a very high solidity factor, e.g. 0.8~0.9, as used in high bypass-ratio turbofans, does not equalize fan wake, and that equalization happens only due to shear friction between the infinitesimal pockets of air themselves.

Found this CFD of a turbofan's fan.

What if there is zero tip clearance, because the shroud is attached to the blades and rotates with them? And you can combine both cases and consider a pipe with rotating section with blades in it...

@JanHudec then the wing and shroud should be seen as a wing.

There is atmospheric pressure outside the wall and higher pressure inside the wall, of course there is going to be flow around the wall. If it were only within the tip clearance then induced drag would limit to zero as tip clearance limits to zero but it doesn't, it limits to the induced drag on an elliptical wing. As to your second claim that downwash velocity isn't constant, we are talking about a perfect irrotational fluid going through an actuator disk.


text:
What is the origin of the ‘six pack’ flight instrument arrangement?
I was wondering who came up with the modern ‘six pack’ flight instrument arrangement, when I realized that the old British blind instrument flying panel that was the standard across many airplane types in the 1940’s was very similar to the ‘six pack’. Is there a direct connection here with the Brits, because at least American aircraft of the era had instrument panels that appeared absurdly disorganized extending into the 1950’s? IOW, the British seemed to have a better idea of where flight instruments were best suited earlier on.

Before the six pack there was the "T"...

The T is part of the six pack.  I believe there was an FAA research project in the early 60s that identified the ideal arrangement for efficient scanning, and GA started to adopt it in the mid 60s. You can see the switch from scattergun arrangements to slightly more organized in the late 50s early 60s then the  T-six pack appears mid 60s.

Thanks. Yes, I remember the "T" now.


text:
How can I pay for flight school in the UK?
I want to apply for the British Airways future pilot programme. However, it costs £125,000. The biggest loan I can get is £20,000 so where am I supposed to get the rest of that money?
I've heard people telling me that you can get a sponsorship, but I've looked online and cannot find any sponsorship deals because it's all self funded.

What level of training / certification do you have so far?  Its one thing for someone with no flight experience at all the aim for B.A.   Its quite different for someone with a PPL / IFR rating to decide they want to aim for an ATP rating.

I have a couple of hours in the 172 but no license

Before you look at a massive loan to become a commercial pilot, perhaps you should make some small steps by getting initial ratings at a more reasonable cost?  Once you have demonstrated exceptional skills at a lower level, you will be more likely to find resources to move up to the next level.  Right now, you're trying to run a marathon, but have never jogged around the block before.

There are specialist firms providing loans for aviation training, however it is unlikely that you will be able to get a loan of the amount you would need for an integrated course without some form of security.
The security usually takes one of two forms.  The first is a charge on a property (belonging to you or your parents).  The other is an airline agreeing to act as a guarantor/sponsor, however this doesn't appear to apply in this case as this particular scheme states it is fully self-funded.
If you cannot raise the financing for this particular scheme, you can apply to other schemes that provide some sponsorship or underwrite the loan as they become available.  Naturally there tends to be more competition for these schemes.  Some are targeted towards under-represented potential pilots, such as the Amy Johnson initiative by Easyjet, where they will guarantee/underwrite the loan for a limited number of female trainees.
It is also possible to undertake modular training, where the cost can be closer to £50,000.  There are a number of organisations that can provide funding towards some of the costs, particularly for young pilots.  These include The Honourable Company of Air Pilots, The Air League and The Royal Aeroclub Trust.
Before spending significant sums on training, it is advisable to get a class one medical, to avoid possible disappointment after spending significant sums of money that cannot be recovered.

Thank you so much for the reply


text:
What is the vertical rod on this Swissair DC-3 drawing?
What is this large protruding rod in the frontal dorsal section of this drawing of a Swissair DC-3:

The small rod extending from the nose I am assuming holds the pitot tube and serves as a mount for underbelly antennas. The other, larger rod I did not find present in cutaway drawings of the DC-3.

I'm not sure that's a DC-3, it's front section is flat sided, the DC-3 is very round.

I think it's a DC-2, it's still an interesting question though

It is definitely a Douglas DC-2 and the large “rod” looks like a radio antenna mast.

It's a defensive machinegun - the Swiss take their armed neutrality very seriously.  ;-P

This is a DC-2 poster from Swissair, circa 1935, a variant can be found here too. The front rod is the pitot support, the rear rod is the sense antenna of the ADF (a long wire running at the bottom of the fuselage, coupled with the loop on the top of the cockpit). wires are not visible though and it leaves an impression of the rod spraying something (this poster is a collector item for conspiracists), maybe addictive cocoa powder:

Source, annotated
They are sometimes closer or grouped:

Source
On Douglas aircraft, later the ADF loop would come on the nose. The HF (radio) antenna was generally found on top of the fuselage, near the tail. All that to say this varied a lot and it's actually difficult to tell what was the antenna purpose if the details are not visible:

Source

Do you mean coca powder?

@phoog: Hey!  This is Switzerland, not Colombia!  We are strictly above-board here!  ( ;-P )

@Vikki hm, I thought the addictive ingredient in chocolate was the sugar rather than the cocoa.


text:
Why do some helicopters have such complicated flight control mixers?
Image source
Some helicopters have a very complicated set-up that mixes flight control inputs to many or all of the swashplate actuators, like in the picture above for the S-76.
Why are all the cross-couplings required, would the helicopter not be controllable without the flight control mixing action? We only need to cant the swashplate for cyclic control, lift it for collective, and connect the pedals to the tail rotor, right?

I don't understand where this question is coming from, so I'm leaning toward the trivial answer "because helicopter flight dynamics are complicated and so is transmitting control inputs to a set of rotating wings". Could you expand the question further?

If you are specifically referencing the image in your question, are you taking into account that most of the complexity comes form the optional servos? If you replace them with rods the entire assy. becomes much clearer. To be honest I can't read what connects to what in that resolution, but it does not seem to do anything out of the ordinary; the only mixing going on is collective input into both the forward and lateral actuators, which is expected, since it is easier to implement collective using the cyclic system already there.

Flight dynamics of a helicopter are complicated indeed. Just to illustrate, consider a helicopter in a stable hover, with the pilot holding the collective, cyclic and pedals in the position that maintains attitude, altitude and heading.
They now want to gain altitude only, in a climbing hover. The sequence of events:

Pilot pulls on the collective, all rotor blades increase angle of attack and the rotor creates more lift. The helicopter climbs.
But because the AoA of the blades have changed, the engine must apply more torque to the rotor and the heading starts to change. So the pilot must apply pedal to change tail rotor thrust and compensate for the torque.
The increased thrust of the tail rotor causes lateral drift. So the pilot needs to input lateral cyclic in order to compensate.

To gain a pure vertical climb, the pilot needs to carry out simultaneous deflections of three flight controls! This collective-to-yaw and collective-to-lateral coupling not only occurs in the hover, but also in forward flight. When increasing speed, the pilot needs to:

deflect cyclic pitch forward;
increase collective for the additional required power;
deflect the pedal for the increase in rotor thrust - but at speed, the vertical fin and the tail rotor take over most or all of this;
deflect the cyclic to the side to compensate for the increased tail rotor thrust, or for the sideslip from the vertical tail stabilisation.

For small, simple and low cost helicopters, this is indeed the way they are flown. But for larger helicopters with a mission, some of the cross couplings are taken care of by the mechanical mixer: when increasing collective, a mechanical link automatically increases the inputs for the cross couplings. It does so in series with the flight controls, which do not move and retain their former selected position.

The drawing in the OP is of at the S-76, a medium sized helicopter, and we can find the following:

Collective/Roll mixer (lateral lead). Provides automatic rotor sideways deflection as a function of collective stick deflection.
Pitch bias actuator. Extends as a function of airspeed, so that the cyclic pitch stays near centre, for more comfort at high speed.
An unlabelled connection rod between collective and pedals.
Collective/Pitch mixer. This helicopter will start to pick up speed when collective is deflected, for a straight-up vertical speed, the pilot will need to deflect the stick backwards.

Also in the OP drawing: optional AFCS servos, which reduce control forces and provide the static stability that helicopter aerodynamics do not provide. Plus trim actuators, which change the neutral position of the feel springs or completely eliminates feel forces. And a yaw damper.
Very complicated indeed.

Not sure if this should be an answer or comment. Not addressing "Mixing" but control system complexity. Many larger turbine helicopters have a system called Force Trim which is used to recenter the cyclic at different speeds. Without Force Trim the pilot would be holding extreme forward cyclic at high speed and extreme aft cyclic at low speeds. Also many models of helicopter have various linkages to minimize feedback from the rotor and control system to the cyclic. The pesky problem of "Stick-shake" can be largely mitigated by some form of irreversible linkage. Both of these systems add to control system complexity.

It's necessary to allow control by the relatively crude "pilot's brain" of an autopilot system, basically to help it out with all the cross-couple responses that are normally taken care of by a human with lots and lots of practice.
You can see that there are autopilot servos in the system, as required for IFR single pilot certification.


text:
Is it possible for a plane to accelerate easily from 30 to 45 mph in flight while it struggles to reach 30 mph (the engine runs at constant power)?
In two letters sent to Dr. G. A. Spratt (an aviation enthusiast), Wilbur Wright reported the progress he and his brother had made, up to September 10, 1904, regarding Flyer II, a plane powered by a 16-17 hp engine. (The pilot had no means to control the motor in flight. He could just stop it, in which case the machine landed.)
The plane needed an airspeed of about 30 mph to fly well. The machine lifted at 23-24 mph (airspeed) but it always fell unless the airspeed reached 25 mph. Between 25 and 30 mph it struggled to keep flying but if somehow reached 30 mph then it accelerated without problem, by itself, to 45 mph.
Assuming the aerodynamic resistance was $$\text{Drag} = kV^2$$
where k is a constant and V was the airspeed, it results that the plane needed the power $$P = \text{Drag} \times V = kV^3$$ to fly at a constant speed, V, and in consequence the necessary power for 45 mph would have been $(45/30)^3 = 3.375$ times greater than that required for flying at 30 mph. In other words, the apparatus would have needed at least 3.375 x 16 hp = 54 hp to reach 45 mph. The difference in power is enormous.
Between 30 and 45 mph the drag should have followed a significantly different law from $\text{Drag} = kV^2$. What would that law be?
1904-08-16, Wilbur Wright, “Letter to G. A. Spratt”, Dayton, August 16, 1904.

Dear Dr. Spratt, ...
Up to Aug. 1st we had made but fourteen trials;
since then we have made seventeen more. So far our longest flight is
only 1304 ft. in 39-1/2 seconds, which though farther over the ground
than our longest flight at Kitty Hawk is not its equal in duration of
time or distance through the air. We are however working under much
less favorable conditions so far as grounds and atmospheric conditions
are concerned. We have found difficulty in getting satisfactory starts
owing to the fact that the winds are usually very light by spells, and
the new machine requires a higher relative speed for starting than the
old one. It lifts at a speed of 23 or 24 miles an hour but the angle
is so great that the resistance exceeds the thrust and the machine
soon stalls. Between 25 and 30 miles it is a case of nip and tuck
between them; but after the relative speed reaches 30 miles, the
thrust exceeds the resistance and the speed accelerates till a
velocity of forty five or fifty miles is reached. So far the highest
speed attained is forty five miles, but it may exceed this when we get
to making longer flights. We are proceeding very cautiously; and do
not intend to attempt anything spectacular until we know that it is
safe, and we know all of the machines peculiar tricks. ... Yours
truly, Wilbur Wright.

1904-09-10, Wilbur Wright, “Letter to G. A. Spratt”, Dayton, September 10, 1904.

Dear Dr. Spratt, ...
We have made forty five starts with our 1904 Flyer. Unless the
relative speed at starting is 27 miles in a calm and two or three
miles more than that in a wind, the machine will gradually slow down
till unable to fly. After the relative speed passes thirty miles the
velocity accelerates till a relative speed of 45 to fifty miles is
reached. We found it difficult in practice to get a speed down the
track greater than 20 miles an hour, so that unless we had a wind of
about 10 miles we were not sure of being able to fly for a lull in the
wind would let us drop below the real flying limit. As we were not
ready to turn our backs to such a wind on account of the enormous
speed in landing when going with the wind our flights have been
confined to the length of our pasture field. We have made a number of
flights between 1250 and 1450 ft long. We have now finished a starting
apparatus which gives a speed at start of 27 miles an hour in a dead
calm, and expect shortly to begin circling. With longer flights and
less hauling the machine back, we hope to get more practice than
heretofore.  With kind regards, Yours truly, Wilbur Wright.

This is just an illustration that shows Flyer I and II. It has nothing to do with the two letters quoted above.

The 1904 Wright Flyer II was almost identical to the 1903 Flyer, as you can see by comparing these two photos. The picture with Flyer I (1903) was first published in “The Wright Brothers’ Aeroplane”, The Century Magazine, New York, September 1908, Vol. LXXVI, No. 5, pp. 641-650. The one with Flyer II (1904) later.

@Federico, I included the illustration to make it clear that I refer to Flyer II of the Wright brothers.

What W. Wright describes in his two letters to G. A. Spratt is pure fiction, imaginary flights.
Using the zero lift drag coefficient taken from the lift and drag diagram of the Wright 1903 machine (a valid approximation as Flyer I and II were quite similar) and also the technical characteristics of Flyer II as it was before September 10, 1904, (without weights of steel added), I reached the conclusion (see the Mathcad worksheet below) that, for the interval of airspeeds ranging between 30 and 45 mph, the drag, when Lift = Weight, increased continuously.
As can be seen in the diagram, the thrust at constant power was always above the total drag till somewhere between 30 and 35 mph but it dropped below after that and in consequence the airplane could not have reached 45 mph.
In theory the plane had enough power to take off and fly without problems only below 35 mph. The story of Wilbur with the rapid acceleration of his apparatus after reaching 30 mph is just just imagination based on wrong beliefs.

The Drag and Thrust of Flyer II (1904). The induced drag formula was taken from here. I also took into account the fact the plane had two wings, one above the other.

the Wrights were very smart, practical people who made extremely accurate observations on the performance of their aircraft.  They would have no reason to engage in "fiction" as they were risking their lives.  What they were describing is consistent with lower induced drag at higher airspeeds.  Their issue may have been with prop pitch.  They had a wind tunnel, and could have tested thrust at various airspeeds.  As you properly mentioned, they also may have had airflow issues over the lower wing, rendering it ineffective at lower airspeeds.

For Flyer II, an airspeed of 40 mph is only possible with about a 30 hp engine and 45 mph with 40 hp. The 1904 motor of 17 hp was too weak even for 35 mph.

What the heck......?

Where do you get the 66% propeller efficiency ?

@MaximEck The propellers of Flyer I (the same as for Flyer II) were "remarkably efficient, converting 66 percent of the mechanical energy from the motor into thrust" (see: http://www.wright-brothers.org/Information_Desk/Just_the_Facts/Engines_&_Props/1903_Props.htm#:~:text=When%20the%20brothers%20finally%20did,from%20the%20motor%20into%20thrust.)

At what airspeed ? Fixed pitch propeller have a huge variation in efficiency relative to wind speed !! If it's 66% staying still that's pretty impressive yes and that efficency will go up to 80% easily when the aircraft is speeding up

@MaximEck The efficiency was 66% when the plane ran at an airspeed of 24 mph (see:https://www.loc.gov/resource/mwright.01006/?sp=6).

Does your mathcadding take into account the ground effect?

Yes and in this case this is due to induced drag.
This drag can be derived with the lift and drag equations of a wing :
$$\textit{Lift} = \frac{1}{2} \rho C_L S V^2$$
$$\textit{Drag} = \frac{1}{2} \rho C_D S V^2$$
With decomposition on first order of the drag coefficient as follow $$C_D = C_{D_0}+k C_L^2$$
We get, $$D = \frac{1}{2} \rho (C_{D_0}+k C_L^2) S V^2 = \frac{1}{2} \rho C_{D_0} S V^2 + \frac{1}{2} \rho S \bigg(k\bigg(\frac{2L}{\rho SV^2}\bigg)^2\bigg)V^2$$
As you can see the second part called induced drag is decreasing with respect of $V^{-2}$. Therefore while flying slow the the faster you go the less drag you will have to overcome as you can see in the following picture. And at one point the other drag forces in the first term, namely pressure drag and friction drag will push back the drag to increase as $V^2$:

As you can see your assumption that the airplane drag increases with $v^2$ is not entirely valid. This works under the assumption of level flight, meaning that your lift remains constant and the induced drag of the wing is therefore only dependent on the air speed. As the wing efficiency decreases with AOA increasing, flying at slow speed and high AOA will generate more induced drag than flying faster with smaller AOA.
Stall of the propeller at low speed might be another explanation why the acceleration is slower. With fixed pitch, thin propeller blades you might enter stall region with to high AOA which will prevent you to have good acceleration at low speed. In the early days of VP props people used to run off the end of runways through trying to take off in coarse pitch. They propeller was stalled and only a small amount of power was transferred to thrust.

@MaximEck The plane was not capable of taking off itself in a calm only because the length of the rail was too short, and could not be extended for technical reasons. W. Wright says in one of the letters that it was difficult for the plane to reach more than 20 mph (in still air) while running on the rail but, with a headwind of 10 mph, the airspeed rose to 30 mph at take-off. In consequence the plane was able to reach an airspeed of 30 mph in a calm assuming the rail would have been long enough.

Okay I have seen design that had that flaw, but in this case it seems to implies that only the inuced drag is the consequence of this behaviour

@MaximEck well, their runway turned out to not be long enough, in either case.

I used your formula for the particular case of Flyer II (https://en.wikipedia.org/wiki/Wright_Flyer_II) and I always get the minimum of the total drag before the interval 30 mph - 45 mph. Besides this, immediately after 30 mph the thrust gets quite close to the total drag and at 35 mph, or even less, the thrust equals the total drag. In other words, the descending part of the total drag is outside the interval 30 - 45 mph of interest.

How did you evaluate the constants $C_{D_0}$ and $k$?

@MaximEck the part about the propeller is very good reasoning, as for L/D, indused drag etc (I admit this is somewhat outside my competence), have you considered the fact that W bros were flying in ground effect? This will have major effects on the dynamics of flight.

To add to prevous, might it be so that flying in  ground effect "extends" or flattens the bottom part of the backside of the power curve? It seems to me the speed region in which W bros were struggling was relatively large.

Another factor could be that the engine produced more power at higher RPM, and flying faster unloaded the engine slightly, allowing it to run a little faster and reach it's power band. Admittedly, this is unlikely for the Wright Flyer, but it's a common issue with RC pylon racers running tuned pipes on 2-stroke engines.

@Jpe61 ground effect actually lower induced drag by decreasing downwash and wingtip vortices. It probably helped them to fly with an underpowered airplane.

This is what i thought. Without ground effect, and the propeller being stalled, they would have been way "deeper" in the backside of the power curve, travelling downwards. With correct technique and given enough altitude, they would have gotten extra speed by diving. Then again, placing the rails on the edge of a cliff might not have been so tempting with such an experimental craft. And it would have been cheating

@MaximEck Cd0 (the zero lift drag coefficient) = 0.08 (see: https://commons.wikimedia.org/wiki/File:Lift_and_Drag_for_Wright_1903_Flyer.gif). With the relations from here https://en.wikipedia.org/wiki/Lift-induced_drag and knowing that the wingspan of Flyer II was 40 ft you can calculate k.

@RobertWerner I answered this question hopefully I know how to use the formulas... My problem is that I had a bit of time on myself, did the calculations and oops it might actually work.

It says here: https://www.uh.edu/engines/epi1867.htm that the W bros tested propellers with a scale model. I wonder how familiar they were with how airflow scales, I'm thinking reynolds number... Did they possibly get the calculations wrong because they used a scalel model for determining thrust? And possibly ended up using wrong prop profile?

@Simplex11 I quickly did the calculation: thrust drag equilibrium around 36mph and a minimal drag value around 28mph assuming perfect propeller efficiency. What we don't know is how this efficiency change with airspeed and ground effect performances. At this minimum drag value they could see this acceleration. But what we also know is that they claim their longest flight 1304 ft. in 39-1/2, that's around 22mph in average, below the written stall speed. So something is off for sure. The behavior I explained there is still valid, whether they measured speed correctly is up to you to decide ;)

"Longest flight 1304 ft. in 39-1/2" into a 10 mph wind?

Yes, and it occurs with nearly every airplane there is.
It is caused by lift-induced drag.

Lift is created by a wing deflecting airflow downward, itself being pushed upward as a result. (Newton's third law)
Lift-induced drag is the backward component of the resulting reaction force acting on a wing. Since at lower speed, the wing needs to be tilted up more to create the necessary lift from the tiny available airfow, the total drag force created by the wing can increase - dramatically.
The Concorde, for example, had a lift-to-drag ratio of 4:1 at takeoff, but 7:1 at Mach 2.
Induced drag follows the inverse of the drag equation you mentioned - it decreases with the square of velocity.
The drag referred to in your equation is parasitic drag which is created by air interacting with the airplane in useless ways, such as squashing against the nose, creating friction on the skin, etc.

But the concorde had a L/D of 12 at M0.95, so it's not always increasing

@MaximEck ah, that's probably parasitic drag

Parasitic, supersonic shock waves, there are plenty of reason but the concorde is not really the best example, as you have nose dropped during take off, and a wing designed to fly supersonic it has very different characteristics than an airliner. In most recent glider for example best glide ratio is reach close to stall speed, just before the inflection of the lift curve with AOA.

@MaximEck I disagree. The basic principle - Newton's third law - remains the same during both flight phases. The Concorde's wing allowed it to fly at very high AoA without stalling, actually simplifying comparison of induced drag at various AoAs (and thus speeds). Moreover, the existence of wave drag at Mach 2 actually serves to make the comparison even more meaningful.

You cannot explain lift and drag using Newton third law !!! Just saw your explanation of lift, it's much more complex than that and has more to do with pressure field than action-reaction. Considering the high AOA of the Concorde, it's just the working principle of the Delta wing. It's stalled but it still generate lift. Your leading edge creates a huge vortice on the upper side of the wing, reducing pressure and generating lift, but it has enormous drag, requiring the postcombustion used on the Concorde

@MaximEck Oh yes we can. Pressure is the manifestation of Newton's third law. Remember, pressure is force/area? Also, that vortex was what I was referring to. Also, why is it a stall? A stall, AFAIK, is when the slope of the lift curve is negative.

How do you explain low pressure on the succion side with Newton 3rd law? Try to calculate how much force is needed to deflect the air by changing its momentum, nowhere near the force required to fly the airplane. Depends on how you define stall, some says when lift decrease, other when lift to drag ratio dramatically increases or even when you have flow separation. The latter two are correct with high AOA delta wings. On traditional wings all 3 arise roughly at the same time

@MaximEck Would that suction exist if the air wasn't being sucked?

Sucked by what ?

Let us continue this discussion in chat.


text:
What date and time definition is used for the second leg of a flight?
Let's take flight BA0015, from London to Syndney, via Singapore.
This is data from FlightStats.
The first leg has this data:

  "carrierFsCode":"BA",
  "flightNumber":"15",
  "departureAirportFsCode":"LHR",
  "arrivalAirportFsCode":"SIN",
  "departureDate":{
    "dateUtc":"2019-08-17T20:30:00Z",
    "dateLocal":"2019-08-17T21:30:00"
  },
  "arrivalDate":{
    "dateUtc":"2019-08-18T09:45:00Z",
    "dateLocal":"2019-08-18T17:45:00"
  },

It takes off from London, on 2019-08-17 at 20:30 UTC
and lands in Singapore, on 2019-08-18 at 09:45 UTC
but then the second leg:

  "carrierFsCode":"BA",
  "flightNumber":"15",
  "departureAirportFsCode":"SIN",
  "arrivalAirportFsCode":"SYD",
  "departureDate":{
    "dateUtc":"2019-08-17T11:25:00Z",
    "dateLocal":"2019-08-17T19:25:00"
  },
  "arrivalDate":{
    "dateUtc":"2019-08-17T19:10:00Z",
    "dateLocal":"2019-08-18T05:10:00"
  },

It takes off, in the past, from Singapore on 2019-08-17 at 11:25 and lands in Sydney on 2019-08-17 at 19:10.
Is is possible that the day doesn't get incremented because the second leg still belongs to the same flight code?

At a guess, it looks like you've asked it to show you the BA15 flight on August 17. It just so happens that there are 2 BA15 departures each day, the earliest (in GMT) is from Singapore and the later one is from London. If you asked for it to display departures on the 18th you would see the Singapore departure, linked to the first flight you've posted here. Is that what your question is about?

You are looking at two different planes.
This is the flight from London to Singapore, which departed on 17th of August, on flightaware.com:

The dates and times are local, meaning the flight departed 2019-08-17T21:11:00Z (UTC) and landed 2019-08-18T10:23:00Z.
Before this flight even took off, another 777 departed Singapore headed for Sydney:

The flight departed 2019-08-17T11:56:00Z and landed 2019-08-17T19:09:00Z.
The plane which landed in Singapore on the 18th of August also continued to Sydney later that day:

ok, so on that day there is the departure flight (LHX->SIN), and the arrival of the previous flight (SIN->SYD), while I was expecting the API to return me one flight LHX->SIN->SYD with its 2 legs. I see now, using the connecting Ids I didn't include in my question that this is indeed two flights.
I was under the assumption that there wouldn't be more than a single flight in a day window (00:00-23:59) flying under the same code, but in this case we can see there are 2: leg1 from the flight of that day, and leg2 from the flight of the previous day.

@Thomas Exactly, they are allowed to use the same flight number multiple times per day, although the ATC callsigns must be unique (see this question).


text:
Why are min and max temperatures (TN TX) not published in TAFs in the United States?
I know that other countries like Brazil, Argentina, etc. do publish minimum and maximum temperatures in their Terminal Aerodrome Forecasts, abbreviated TN and TX. But why are they not shown in TAFs in the United States? (I guess other countries in Europe also have the temperatures).
BTW, where are the expected min and max temperatures in the US published in an official report? NOAA/NWS? WX-BRIEF?

This isn't much of an explanation, but the National Weather Service says:

The international TAF also contains forecast temperature, icing, and
  turbulence. These three elements are not included in National Weather
  Service (NWS) prepared TAFs. The U.S. has no requirement to forecast
  temperatures in an aerodrome forecast

That sounds to me like "we don't think it's worth it". There are various temperature forecasts on the NWS aviation site (icing level, surface temperatures, graphical area forecast) but personally speaking, I get low/high temperature information from regular, non-aviation forecasts. You can get those forecasts from the NWS site too, if that's what you mean by "official".

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How to log Flight Training Device (FTD) and Full Flight Simulator (FFS) time in the US?
I just completed the ATP CTP Course in the US as a pre-requisite for taking the ATM Written Exam. During the course, according to the syllabus, I have flown 4 hours of B733 in a FTD (fixed-base) and 6 more hours of B733 in a FFS (motion-base). The Simulator is FAA Certified. How should I log this pilot time in my logbook, according to the US Regulations? Do this simulator time counts as total time?

See this question for information on "total time". It isn't defined by regulation. I believe airlines sometimes go into more detail than the regulations in how time is logged, e.g. they may distinguish between PIC time from acting as PIC vs. PIC time from being the sole manipulator of the controls. Good general advice for logging is to use an electronic logbook and log as much detail as possible about each flight or sim session. Then you can easily find time that meets whatever criteria you may be asked for.

Your flight instructor should be the one to answer that, and maybe even write the first entry for you as an example of how to log the time.


text:
What is the difference between Marginal VFR (MVFR) and Special VFR (SVFR)?
I would like to know the difference between marginal VFR (MVFR) and Special VFR (SVFR), particularly when flying in the US airspace. What does each of them mean exactly?

You may be confusing two things. MVFR is a weather condition, Special VFR is a type of operation. You asked what they mean.

Marginal VFR
Straight out of the Aviation Weather Services advisory circular (AC 00-45G with changes 1 and 2):

Marginal Visual Flight Rules (MVFR) indicated on the Weather Depiction
Chart represents ceiling 1,000 to 3,000 feet and/or visibility 3 to 5
statute miles and VFR operations can take place. MVFR areas are
outlined with a solid line, but the area is not shaded. MVFR areas are
typically shaded blue in colorized versions of the chart.

For all regulatory purposes, this has no rule changes from VFR.  I believe the reason it exists is for pilots to more easily identify areas where the weather conditions are approaching IMC.
Special VFR in the US
Special VFR allows a pilot to fly within the lateral boundaries of controlled airspace to the ground with at least 1SM visibility and clear of clouds.
Special VFR is something that can be granted by ATC upon request by a pilot.  It may only be granted within the lateral boundaries of controlled airspace that goes to the ground.
If the special VFR request is being made at night, the pilot and airplane must be IFR legal.
JO 7110.65Y Section Regarding Special VFR
There is no requirement that the weather conditions being reported are below basic VFR.

The Difference
MVFR is an advisory term.
Special VFR is an, on request by pilot only, "required weather minimums" change that may be approved by ATC.
For the purposes of special VFR in the US, here is the FAA regulation. (14 CFR 91.157)

Special VFR exists across the world: http://www.skybrary.aero/index.php/Special_VFR. I've never heard "marginal VFR" being used as an official term, though, so I'm guessing that's FAA specific.

Just curious, where exactly in the FARs does it say that the airport must be reporting IFR conditions at the surface for Special VFR to be granted?  I'm imagining a case where the conditions are just barely above IFR-- say, there is an overcast layer at 1100' AGL-- and a pilot wants to request Special VFR as a precaution in case the conditions deteriorate while he is still in the airspace in question.

Also, there seems to be some ambiguity as to whether not Special VFR may be granted in E4 Class-E-to-Surface "extensions".  For more, see related answer https://aviation.stackexchange.com/a/75022/34686, from the beginning of the answer through the end of the quote "Our opinion is that E4 airspace is not..."

@quietflyer I think the matter of class E is referring to those little squared-off pieces of class E for the purpose of protecting instrument approaches that step outside of the primary airspace ring for the airport.

@quietflyer I did more research, and it turns out the reason I believed that to be true was a bad lead.  I had a TRACON once deny SVFR because the weather minima was not IFR at the surface, but it looks like maybe that controller just made a mistake.

From an ATC perspective, MVFR conditions (ceiling 5000 or below, visibility 5SM or below) are one of the triggers for soliciting PIREPs. But that isn't a regulatory thing, just part of our orders.

In one sentence, then, MVFR is a description of weather conditions, while SVFR refers to a type of operation, yes?

source: pilotinstitute.com
To the right side on this chart is ceilings. No matter the visibility can still be MVFR, IFR or low IFR conditions. SVFR clearance can be given at or above but not below 1sm and is required if ceilings are at or below 3000ft

Welcome to aviation.stackexchange.com!  I've added a cropped version of the screenshot you posted as a second answer to this answer and removed the other answer. Please note it is possible to edit answers and questions after you have posted them. If you have some time, please take the tour to see how this website and its community work.

Re "SVFR clearance ... is required if ceilings are at or below 3000ft"  -- umm, 1) I assume that "SVFR clearance .. is required" is shorthand for "VFR operations without a SVFR clearance are prohibited".   2) shouldn't that be 1000' not 3000'?  3)  shouldn't that only apply to airspace that is "beneath (a) ceiling... within the lateral boundaries of controlled airspace designated to the surface for an airport"?  I.e., for example, not in Class G airspace! See FAR 91.155(c) https://www.law.cornell.edu/cfr/text/14/91.155

The OP doesn't ask for weather differences. He asks what they mean.


text:
What is the difference between Distant (D) NOTAMs, Local (L) NOTAMs and FDC NOTAMs in the US?
I would like to know the difference between Distant (D) NOTAMs, Local (L) NOTAMs and FDC NOTAMs in the US. Is their meaning obsolete or still in use by the FAA?

Short answer: D NOTAMs are about airports and navigation facilities; FDC NOTAMs are about airspace, procedures or regulations; and L NOTAMs don't exist any more (unless you're in the military).

The information on NOTAM types is a bit messy and inconsistent, to say the least. The Alaskan FSS lists 11 types; the PHAK 13-3 lists 3 types (as acpilot mentioned in his answer); the AIM 5-1-3 lists five in its NOTAM section, plus GPS NOTAMs in the GPS section.
And the FAA's military NOTAM site lists yet more types in their FAQ  such as M, V, W and F.
Anyway, here's what the AIM says about D and FDC NOTAMs:

NOTAM (D) information is disseminated for all navigational facilities
  that are part of the National Airspace System (NAS), all public use
  airports, seaplane bases, and heliports listed in the Chart Supplement
  U.S. [...]  NOTAM (D) information includes such data as taxiway
  closures, personnel and equipment near or crossing runways, and
  airport lighting aids that do not affect instrument approach criteria,
  such as VASI.
[...]
FDC NOTAMs. On those occasions when it becomes necessary to
  disseminate information which is regulatory in nature, the National
  Flight Data Center (NFDC), in Washington, DC, will issue an FDC NOTAM.
  FDC NOTAMs contain such things as amendments to published IAPs and
  other current aeronautical charts. They are also used to advertise
  temporary flight restrictions caused by such things as natural
  disasters or large-scale public events that may generate a congestion
  of air traffic over a site.

According to the Alaska FSS page, NOTAM(L) doesn't exist any more for civilian use:

NOTAM (L):  A NOTAM given local dissemination by voice and other
  means, such as telautograph and telephone, to satisfy local user
  requirements.  This type of NOTAM is now used only in the military
  NOTAM system.  All NOTAMs previously considered NOTAM (L)s in the
  civil NOTAM system are now considered NOTAM (D)s.

That's confirmed by AIR FORCE INSTRUCTION 11-208 (I), which describes NOTAM(L) as a military type:

4.2.1. Military.
  [...]
4.2.1.2. Local NOTAM (L series). Local NOTAMs are non-critical and non-safety related information that is essential for the flying and
  support communities to know. Facility managers may establish the
  parameters for local NOTAM determination.

Civilian L NOTAMs are now D NOTAMs that start with the specific facility code. You can find them on the civilian NOTAM site, e.g.:

!LOU 05/077 LOU RWY 15/33 CLSD
  1605091213-1606222000

And to address your final point, NOTAMs are not obsolete. There needs to be some way to inform pilots about temporary changes or things that happen at short notice, and that's what NOTAMs are for. Many people feel that the system could be improved somehow - pulling up the NOTAMs for even a short flight can give you dozens of results - but there's no doubt that it's needed. Reviewing NOTAMs is an important preflight step, and examiners often test you on it during checkrides. There are plenty of NTSB accident reports that mention NOTAMs, usually in cases where the pilot didn't review them.

+1 for stating that L is not a thing for civilians anymore.

A really great source for these kinds of answers is the Pilot's Handbook of Aeronautical Knowledge, or PHAK.  You can dowload it for free from the FAA here.
The answer to your question from chapter 13, pages 13-3 & 13-4 of the PHAK:

NOTAMs provide the most current information available.  They provide
time-critical information on airports and  changes that affect the
national airspace system (NAS) and are of concern to IFR operations.
NOTAM information is classified into three categories. These are
NOTAM-D or  distant, NOTAM-L or local, and flight data center (FDC)
NOTAMs. NOTAM-Ds are attached to hourly weather reports and are
available at automated flight service stations (AFSS) or FSS.
FDC NOTAMs are issued by the National Flight Data Center and contain
regulatory information, such as temporary flight restrictions or an
amendment to instrument approach procedures. The NOTAM-Ds and FDC
NOTAMs are contained in the NOTAM publication, which is issued every
28 days. Prior to any flight, pilots should check for any NOTAMs that
could affect their intended flight.
NOTAM-D information includes such
data as taxiway closures, personnel and equipment near or crossing
runways,  and airport lighting aids that do not affect instrument
approach criteria, such as visual approach slope indicator (VASI).
NOTAM-D information is distributed locally only and is not attached to
the hourly weather reports. A separate file of local NOTAMs is
maintained at each FSS for facilities  in their area only. NOTAM-D
information for other FSS areas must be specifically requested
directly from the FSS that has responsibility for the airport
concerned.

NOTAMs are far from obsolete.  They are critical and pilots are actually required to familiarize themselves with all applicable information before each flight.  NOTAMs are arguably the most important vehicles to help pilots satisfy that requirement.
The civilian NOTAM-L has been discontinued snd evetything that one qualified as a -L is now -D.

You list good details on the NOTAM-D and (FDC) NOTAM. Is there more info available on the NOTAM-L? If you could include a bit on that, it would make a great answer more complete. Also, I'm confused on why (from para 3) "NOTAM-D information is distributed locally only" - that seems to be a contradiction. Is it a typo or am I just not understanding something?

NOTAM-L will cover details relevant to local operations that likely won't effect pilots at other airports.  Ramp repavement and vegetation removal are a couple of examples.  Runway closures qualify as a -D because that would effect flight planning of pillots from other airports.

@FreeMan Yes, NOTAM-L has been combined with NOTAM-D for civilian use.  Now it is only NOTAM-D and FDC.  This is a relatively new change.

Advisory Circulars are also a great source of information for pilots. "FAA policy changes have resulted in significant changes to the format of Distant (D) and Local (L) NOTAMs:   • Prior civil “L” NOTAMs will be reclassified as “D” NOTAMs (Military L series will remain unchanged"

Just to clarify, that advisory circular above is canceled, but that change is still accurate. This was when it was initially changed in 2008.  All we have now are NOTAM-(D) and FDCs.

If, as @RyanMortensen says, (and Pondlife agrees), NOTAM-L doesn't exist anymore (at least in civilian use), then this answer could be updated to reflect that.

I don't understand the down vote on my answer.  Is there a problem with it or was it a move to push it below the other answer?  It was made after I modified my asnwer to include the changes to NOTAM-L.

authors: владимир шипилов